Find the Mth element of the Array after K left rotations

Given non-negative integers K, M, and an array arr[] with N elements find the Mth element of the array after K left rotations.

Examples:

Input: arr[] = {3, 4, 5, 23}, K = 2, M = 1
Output: 5
Explanation: 
The array after first left rotation a1[ ] = {4, 5, 23, 3}
The array after second left rotation a2[ ] = {5, 23, 3, 4}
st element after 2 left rotations is 5.

Input: arr[] = {1, 2, 3, 4, 5}, K = 3, M = 2
Output:
Explanation: 
The array after 3 left rotation has 5 at its second position.

Naive Approach: The idea is to Perform Left rotation operation K times and then find the Mth element of the final array.



Time Complexity: O(N * K)
Auxiliary Space: O(N)

Efficient Approach: To optimize the problem, observe the following points:

  1.  If the array is rotated N times it returns the initial array again. 
     

    For example, a[ ] = {1, 2, 3, 4, 5}, K=5 then the array after 5 left rotation a5[ ] = {1, 2, 3, 4, 5}. 

    Therefore, the elements in the array after Kth rotation is the same as the element at index K%N in the original array.

  2. The Mth element of the array after K left rotations is
     

    { (K + M – 1) % N }th

    element in the original array.

 

Below is the implementation of the above approach:

C++

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// C++ program for the above approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to return Mth element of 
// array after k left rotations 
int getFirstElement(int a[], int N, 
                    int K, int M) 
    // The array comes to original state 
    // after N rotations 
    K %= N; 
  
    // Mth element after k left rotations 
    // is (K+M-1)%N th element of the 
    // original array 
    int index = (K + M - 1) % N; 
  
    int result = a[index]; 
  
    // Return the result 
    return result; 
  
// Driver Code 
int main() 
    // Array initialization 
    int a[] = { 3, 4, 5, 23 }; 
  
    // Size of the array 
    int N = sizeof(a) / sizeof(a[0]); 
  
    // Given K rotation and Mth element 
    // to be found after K rotation 
    int K = 2, M = 1; 
  
    // Function call 
    cout << getFirstElement(a, N, K, M); 
    return 0; 

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Java

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// Java program for the above approach 
import java.util.*; 
class GFG{ 
      
// Function to return Mth element of 
// array after k left rotations 
public static int getFirstElement(int[] a, int N, 
                                  int K, int M) 
      
    // The array comes to original state 
    // after N rotations 
    K %= N; 
  
    // Mth element after k left rotations 
    // is (K+M-1)%N th element of the 
    // original array 
    int index = (K + M - 1) % N; 
  
    int result = a[index]; 
  
    // Return the result 
    return result; 
  
// Driver code 
public static void main(String[] args) 
      
    // Array initialization 
    int a[] = { 3, 4, 5, 23 }; 
  
    // Size of the array 
    int N = a.length; 
  
    // Given K rotation and Mth element 
    // to be found after K rotation 
    int K = 2, M = 1
  
    // Function call 
    System.out.println(getFirstElement(a, N, K, M)); 
  
// This code is contributed by divyeshrabadiya07 

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Python3

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# Python3 program for the above approach 
  
# Function to return Mth element of 
# array after k left rotations 
def getFirstElement(a, N, K, M): 
      
    # The array comes to original state 
    # after N rotations 
    K %=
  
    # Mth element after k left rotations 
    # is (K+M-1)%N th element of the 
    # original array 
    index = (K + M - 1) %
  
    result = a[index] 
  
    # Return the result 
    return result 
  
# Driver Code 
if __name__ == '__main__'
      
    # Array initialization 
    a = [ 3, 4, 5, 23
  
    # Size of the array 
    N = len(a) 
  
    # Given K rotation and Mth element 
    # to be found after K rotation 
    K = 2
    M = 1
  
    # Function call 
    print(getFirstElement(a, N, K, M)) 
  
# This code is contributed by mohit kumar 29 

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C#

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// C# program for the above approach 
using System;
  
class GFG{
      
// Function to return Mth element of 
// array after k left rotations 
public static int getFirstElement(int[] a, int N, 
                                  int K, int M) 
      
    // The array comes to original state 
    // after N rotations 
    K %= N; 
  
    // Mth element after k left rotations 
    // is (K+M-1)%N th element of the 
    // original array 
    int index = (K + M - 1) % N; 
  
    int result = a[index]; 
  
    // Return the result 
    return result; 
  
// Driver code
public static void Main(string[] args) 
{
      
    // Array initialization 
    int []a = { 3, 4, 5, 23 }; 
  
    // Size of the array 
    int N = a.Length; 
  
    // Given K rotation and Mth element 
    // to be found after K rotation 
    int K = 2, M = 1; 
  
    // Function call 
    Console.Write(getFirstElement(a, N, K, M));
}
}
  
// This code is contributed by rutvik_56

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Output: 

5

 

Time complexity: O(1)
Auxiliary Space: O(1)

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